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Mathematics 11 Online
OpenStudy (anonymous):

|x+1|=2x

OpenStudy (goten77):

kinda obvious what x is but i dont kno how to explain or do this 1 mathmetically

hartnn (hartnn):

if |y|=a then \[y=\pm a\] so |x+1|=2x gives (x+1)= 2x or (x+1)=-2x can u solve these two to get 2 values of x ?? out of these 2 ,one after re substituting in original equation,will not give u equality.so u discard that vale. then u will have only 1 solution.

OpenStudy (anonymous):

@hartnn so how do i do |x+1|=2x

OpenStudy (anonymous):

subtract 1 from both sides? @hartnn

hartnn (hartnn):

as i have explained,u have (x+1)= 2x or (x+1)=-2x u need to solve both of these take one at a time x+1=2x subtract x from both sides,what u get??

OpenStudy (anonymous):

1x? @hartnn

OpenStudy (anonymous):

right?

hartnn (hartnn):

1x where on left side or right side?

OpenStudy (anonymous):

right side @hartnn

hartnn (hartnn):

so whats on left side??

OpenStudy (anonymous):

um is it 1

hartnn (hartnn):

yup,so u have 1x=1 that is x=1 is one of your solution. can u get other solution similarly ?

OpenStudy (anonymous):

Thanks and i'll try

OpenStudy (anonymous):

so the second on is |x+1|=-2x right? @hartnn

hartnn (hartnn):

(x+1)=-2x u can remove mod | | sign,since u are taking -2x

OpenStudy (anonymous):

is x=-3 correct?

OpenStudy (anonymous):

@hartnn

hartnn (hartnn):

nopes, u get 3x=-1 so x=-(1/3) but when u put this in original equation,equality does not get satisfied, so the only solution u have is x=1 ok?

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